The proof of the theorem is not too hard. What we're looking for is
what's called an automorphism of the complex numbers. This is
a function, f, which "relabels" the complex numbers, so that
arithmetic on the new labels is the same as the arithmetic on the old
labels. For example, if 3×4 = 12, then f(3) ×
f(4) should be f(12).

Let's look at a simpler example, and consider just the integers, and
just addition. The set of even integers, under addition, behaves just
like the set of all integers: it has a zero; there's a smallest
positive number (2, whereas it's usually 1) and every number is a
multiple of this smallest positive number, and so on. The function
f in this case is simply f(n) = 2n, and it
does indeed have the property that if a + b = c,
then f(a) + f(b) = f(c) for
all integers a, b, and c.

Another automorphism on the set of integers has g(n) =
-n. This just exchanges negative and positive. As far as
addition is concerned, these are interchangeable. And again, for all
a, b, and c, g(a) +
g(b) = g(c).

In fact, there are no interesting automorphisms on the integers that
preserve both addition and multiplication. To see this, consider an
automorphism f. Since f is an automorphism that
preserves multiplication, f(n) = f(1 × n) =
f(1) × f(n) for all integers n. The
only way this can happen is if f(1) = 1 or if f(n) = 0
for all n.

The latter is clearly uninteresting, and anyway, I neglected to
mention that the definition of automorphism rules out functions that
throw away information, as this one does. Automorphisms must be
reversible. So that leaves only the first possibility, which is that
f(1) = 1. But now consider some positive integer n.
f(n) = f(1 + 1 + ... + 1) = f(1) +
f(1) + ... + f(1) = 1 + 1 + ... + 1 = n. And
similarly for 0 and negative integers. So f is the identity
function.

One can go a little further: there are no interesting automorphisms of
the real numbers that preserve both addition and multiplication. In
fact, there aren't even any reasonable ones that preserve addition.
The proof is similar. First, one shows that f(1) = 1, as
before. Then this extends to a proof that f(n) =
n for all integers n, as before. Then suppose that
a and b are integers.
b·f(a/b) =
f(b)f(a/b) =
f(b·a/b) = f(a) = a,
so f(a/b) = a/b for all rational numbers
a/b. Then if you assume that f is continuous,
you can fill in f(x) = x for the irrational
numbers also.

(Actually this is enough to show that the only continuous
addition-preserving automorphism of the reals is the identity
function. There are discontinuous addition-preserving functions, but
they are very weird. I shouldn't need to drag in the continuity issue
to show that the only addition-and-multiplication-preserving
automorphism is the identity, but it's been a long day and I'm really
fried.)

[ Addendum 20060913: This previous paragraph is entirely wrong; any
function x → kx is an addition-preserving
automorphism, except of course when k=0. For more complete
details, see this later article. ]

But there is an interesting automorphism of the complex numbers; it
has f(a + bi) = a - bi for all real
a and b. (Note that it leaves the real numbers fixed,
as we just showed that it must.) That this function f is an
automorphism is precisely the content of the statement that i
and -i are numerically indistinguishable.

The proof that f is an automorphism is very simple. We need to
show that if f(a + bi) + f(c +
di) = f((a + bi) + (c + di))
for all complex numbers a+bi and c+di, and
similarly f(a + bi) × f(c +
di) = f((a + bi) × (c +
di)). This is really easy; you can grind out the algebra in
about two steps.

What's more interesting is that this is the only nontrivial
automorphism of the complex numbers. The proof of this is also
straightforward, but a little more involved. The purpose of this
article is to present the proof.

Equation 2 implies that either p or q is 0. If they're
both zero, then f(a + bi) = a, which is
not reversible and so not an automorphism.

Trying q=0 renders
equation 1 insoluble because there is no real number p with
p2 = -1. But p=0 gives two solutions.
One has p=0 and q=1, so f(a+bi) =
a+bi, which is the identity function, and not
interesting. The other has p=0 and q=-1, so
f(a+bi) =
a-bi, which is the one we already knew about. But we
now know that there are no others, which is what I wanted to show.